A body falling from a high Minaret travels $40$ meters in the last $2$ seconds of its fall to ground. Height of Minaret in meters is (take $g = 10\,m/{s^2}$)
A body falling from a high Minaret travels $40$ meters in the last $2$ seconds of its fall to ground. Height of Minaret in meters is (take $g = 10\,m/{s^2}$)
$60$
$45$
$80$
$50$
A body falling from a high Minaret travels $40$ meters in the last $2$ seconds of its fall to ground. Height of Minaret in meters is (take $g = 10\,m/{s^2}$)
Let height of minaret is $H$ and body take time $T$ to fall from top to bottom.
$H = \frac{1}{2}g{T^2}$…(i)
In last $2\, sec$. body travels distance of $40\,meter $ so in $(T - 2)\,sec$ distance travelled $ = (H - 40)\;m$.
$(H - 40) = \frac{1}{2}g{(T - 2)^2}$…(ii)
By solving (i) and (ii)
$T = 3\,sec$ and $H = 45\;m$.